Question

AB and AC are two equal chords of circle. prove that bisector of the angle BAC passes through the centre of circle​

Answer 1

Answer: Proved

Step-by-step explanation:

Let the two chords AB and AC are chords

centre of the circle o lies on the bisector of the ∠BAC

Now join BC

let the bisector of ∠ B A C intersect BC at P

Now from ΔAPB and ΔAPC

AB =AC

BAP =CAP

AP=AP

ΔAPB ≅  Δ APC

BP=CP and ∠APB=∠ APC     ( By CPCT)

Now ∠APB + ∠APC = 180° ( linear pair)

2 ∠APB = 180

∠APB = 180 / 2

∠APB = 90°

So. BP =CP  ,

Hence AP is the perpendicular bisector of the chord BC and AP passes through the centre of the circle .

Answer 2

Question:-

AB and AC are two equal chords of circle. prove that bisector of the angle BAC passes through the centre of circle.

Required Answer:-

Given:

AB and CD are two equal chord of the circle.

Prove:

Center O lies on the bisector of the ∠BAC.

Construction:

Join BC. Let the bisector on ∠BAC intersect BC in P.

Proof:

In△APB and △APC,

AB=AC ...(Given)

∠BAP=∠CAP ...(Given)

AP=AP ....(common)

∴△APB≅△APC ...SAS test

⇒BP=CP and ∠APB=∠APC ...CPCT

∠APB+∠APC=180. ...(Linear pair)

∴2∠APB=180 ....(∠APB=∠APC)

⇒∠APB=90

∴AP is perpendicular bisector of chord BC.

Hence, AP passes through the center of the circle.

Brainliest .. !! ⚡❤

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